The spring balance A reads 2 , kg with a bloc | vedclass

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(D) When the block $m$ is immersed in the liquid,it experiences an upward buoyant force (upthrust) $F_B$ according to Archimedes' principle.For balanc

Vedclass · Accepted Answer

Both $(b)$ and $(c)$.

Vedclass · Answer

(D) When the block $m$ is immersed in the liquid,it experiences an upward buoyant force (upthrust) $F_B$ according to Archimedes' principle.For balance $A$,the reading is the tension $T$ in the string. Initially,$T = mg = 2 \, kg \cdot g$. When immersed,$T = mg - F_B$. Since $F_B > 0$,the reading of balance $A$ will be less than $2 \, kg$.For balance $B$,it measures the normal force exerted by the beaker on the pan. Initially,it reads the weight of the beaker plus the liquid. When the block is immersed,the liquid exerts an upward force $F_B$ on the block. By Newton's third law,the block exerts an equal and opposite downward force $F_B$ on the liquid. Thus,the total downward force on the balance $B$ increases by $F_B$. Therefore,the reading of balance $B$ will be more than $5 \, kg$.Since both statements $(b)$ and $(c)$ are correct,the correct option is $(d)$.

The spring balance $A$ reads $2 \, kg$ with a block $m$ suspended from it. $A$ balance $B$ reads $5 \, kg$ when a beaker filled with liquid is put on the pan of the balance. The two balances are now arranged such that the hanging mass is inside the liquid as shown in the figure. In this situation:

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